Let \(X=\omega ^\omega \)  with the (continuous) shift map \(T\) . For \(n\in \omega \) 
write \(\delta _{\bar{n}}\)  for the point-mass measure on the constant \(n\) 
sequence denoted \(\bar{n}\) . Let \(V=\lbrace  \delta _{\bar{0}}\rbrace \) .
To show that \(G(V)\)  is \(\Pi ^1_1\) -complete, we construct a continuous map
\(\pi \colon \text{Tr}\rightarrow X\)  reducing \(\text{Wf}\)  to \(G(V)\) , where \(\text{Wf}\) 
is the set of all wellfounded trees on \(\omega \) . Recall that a tree \(\mathcal {T}\)  on \(\omega \)  is a
set \(\mathcal {T}\subseteq \omega ^{<\omega }\)  of finite length sequences from \(\omega \)  which is
closed under the operation of taking initial segments. A tree is called wellfounded if it doesn't contain an infinite branch, and illfounded if it does. Let \(\text{Tr}\)  denote the set of all trees on \(\omega \) .
We fix an enumeration \(t^{(1)},t^{(2)},\ldots \)  of \(\omega ^{<\omega }\)  such that
\(|t^{(i)}|\le |t^{(j)}|\)  for \(i\le j\) . Using this enumeration we identify
\(\omega ^{<\omega }\)  with \(\omega \)  and \(\text{Tr}\)  with a closed subset of the
compact Polish space \(2^\omega \) .
Thus, \(\text{Tr}\)  is a Polish space and \(\text{Wf}\subseteq \text{Tr}\)  is a \(\Pi ^1_1\) -complete set
in \(\text{Tr}\)  (see [1]}). Note that two trees in \(\text{Tr}\)  are close
if and only if there is \(n\)  such that for each \(1\le i \le n\) , the
finite word \(t^{(i)}\)  either belongs to both trees or belongs to neither of them.
First, we will associate to each finite word \(s\in \omega ^{<\omega }\) 
a measure \(\mu _s\) , which will be a
finite combination of Dirac point-mass measures.
Let us fix some auxiliary notation. By \(p(n)\)  we will denote the \(n\) -th prime
(that is, \(p(0)=2\) , \(p(1)=3\)  etc.). Fix a bijection
\(\mathbf {n}\colon \omega \times \omega \rightarrow \omega \) . Let \(Q\)  be the set of
positive integers divisible by at least two different primes.
Let \(q(n)\)  denote the \(n\) -th smallest element of \(Q\) . That is, \(q(0)=6\) , \(q(1)=10\) , \(q(2)=12\)  etc. Given \((n,i)\in \omega \times \omega \)  we set \(\nu _{n,i}\)  to be the Dirac point-mass
measure concentrated on the constant sequence \(\bar{p}=ppp\ldots \in \omega ^\omega \) 
where \(p=p(\mathbf {n}(n,i))\) .
For \(s=(s(0)\ldots s(\ell -1))\in \omega ^{<\omega }\)  (so \(\ell =|s|\)  is the length of \(s\) )
we define a measure \(\mu _s\)  as
\(\mu _s=\frac{1}{2^\ell }\delta _{\bar{1}}+\sum _{i<\ell } \frac{1}{2^{i+1}}\nu _{s(i),i}.\) 
Suppose \(s_0, s_1, s_2,\dots \)  is a sequence from \(\omega ^{<\omega }\) 
with \(\lim _i |s_i|=\infty \)  and such that \(s_i\)  converges to \(x=(x(0),x(1),\dots ) \in \omega ^\omega \) ,
that is, \(\forall m\ \exists n\ \forall i\ge n\ (s_i\upharpoonright m=x\upharpoonright m)\) .
Then
the corresponding sequence of measures \(\mu _{s_i}\)  is easily seen to converge to the
measure \(\mu _{x}\)  as \(i\rightarrow \infty \) , where \(\mu _x\)  is the measure given by
\(\mu _{x}=\sum _{i=0}^\infty \frac{1}{2^{i+1}}\nu _{x(i),i}=\sum _{i=0}^\infty \frac{1}{2^{i+1}}\nu _{p(\mathbf {n}(x(i),i))}.\) 
On the other hand assume that \(s_0, s_1, s_2,\ldots \)  is a sequence of distinct
elements of \(\omega ^{<\omega }\)  and the corresponding sequence of measures
\(\mu _{s_i}\)  is Cauchy. We claim that \(\lim _i |s_i|=\infty \)  and there is an \(x \in \omega ^\omega \) 
such that \(s_i\)  converges to \(x\) . Let \(k\)  be the least integer, if one exists,
such that either there are infinitely many \(i\)  with \(s_i(k)\)  not defined or
\(\lbrace  s_i(k) \rbrace \)  is not eventually constant. Assuming \(k\)  is defined, the first case cannot
happen as then \(s_i\upharpoonright k\)  is defined and constant for all large enough \(i\)  but
for some \(i<j\)  we would have \(s_i=s_j\) . If the second case
in the definition of \(k\)  occurs, then for large enough \(i\)  we have
that \(s_i\upharpoonright k\)  is constant and either \(s_i(k)\)  takes infinitely many values or
there are two distinct values of \(s_i(k)\)  both of which occur for infinitely
many values of \(i\) .
However, we would then have
\(D(\mu _{s_i}, \mu _{s_j})\ge 1/2^{k+1}\)  for arbitrarily large \(i<j\) , a contradiction.
So, \(k\)  is not defined, that is,
for all \(k\)  we have that for all large enough \(i\)  that \(s_i(k)\)  is defined
and constant. This shows that the \(s_i\)  converge to some \(x \in \omega ^\omega \) .
Finally, for each \(s=(s(0)\ldots s(\ell -1))\in \omega ^{<\omega }\)  and \(0\le i<\ell \) 
we set \(n^s_i=p(\mathbf {n}(s(i),i))\)  and define a periodic point
\(y^s=\left(\underbrace{n^s_0\ldots n^s_0}_{2^{2\ell }\text{ times}}\ \underbrace{n^s_1\ldots n^s_1}_{2^{2\ell -1}\text{ times}}\ \ldots \ \underbrace{n^s_{\ell -1}\ldots n^s_{\ell -1}}_{2^{\ell +2}\text{ times}}\ \underbrace{1\ldots 1\vphantom{n^s_1}}_{2^{\ell +1}\text{ times}}\right)^\infty \in \omega ^\omega .\) 
Note that the primary period of \(y^s\)  equals \(p_s=2^{2\ell +1} - 2^{\ell + 1}\)  and
\(D(\operatorname{\mathcal {E}}(y^s, p_s),\mu _s)\)  goes to 0 as \(\ell \)  goes to infinity.
We are ready to define \(\pi (\mathcal {T})\)  for a tree \(\mathcal {T}\) .
Let \(c_n\)  be the primary period of \(y^s\)  for \(s=t^{(n)}\) .
We fix sequences \(( a_n)\)  and \((b_n)\)  satisfying
\(a_1=2\) , \(a_{2n}=c_n\) , \(a_{2n+1}=1\) , \(b_{n+1} > b_n\)  and
\( a_nb_n> \max \lbrace  2^n a_{n+1}, 2^n (a_0 b_0+\cdots +a_{n-1}b_{n-1})\rbrace \) 
for all \(n\) .
We set \(\ell _0=0\) . As stage \(n\ge 1\)  of the construction of \(x=\pi (\beta )\) 
we will define \(x_{[\ell _{n-1},\ell _n)}\)  where \(\ell _n=\ell _{n-1}+ a_n b_n\) .
For \(n\)  odd we set \(x_{[\ell _{n-1},\ell _n)}={\bar{0}}_{[\ell _{n-1},\ell _n)}=0\ldots 0\) 
(here 0 appears \(a_nb_n\)  times). For \(n\)  even,
we have two cases: if \(s=t^{(n/2)}\in \mathcal {T}\) , then we put \(x_{[\ell _{n-1},\ell _n)}=(y^s)^{b_n}\) .
Otherwise (that is, if
\(s=s^{(n/2)}\notin \mathcal {T}\) ), we put \(x_{[\ell _{n-1},\ell _n)}=(q(n/2))^{a_nb_n}\) .
Note that \(\pi (\mathcal {T})=x\)  defined in this way depends continuously on \(\mathcal {T}\) 
because \(x_{[0,\ell _{2n})}\)  is determined by
\(\lbrace t^{(1)},\ldots ,t^{(n)}\rbrace \cap \mathcal {T}\) . Furthermore, it follows from (REF ) that the
measures \(\operatorname{\mathcal {E}}(x, \ell _{2n-1})\)  converge to \(\delta _{\bar{0}}\)  as \(n\rightarrow \infty \) .
Thus \(\delta _{\bar{0}} \in V(x)\) .
We show that \(\delta _{\bar{0}}\)  is the only measure in \(V(x)\)  if and only if
\(\mathcal {T} \in \text{Tr}\)  is a wellfounded tree.
Let \(\nu \in V(x)\)  and \(i_k\)  be a strictly increasing sequence with
\(\lim _{k\rightarrow \infty } \mathcal {E}(x, i_k)=\nu \) . Let \((n_k)\) 
be a sequence of integers such that \(\ell _{n_k} \le i_k < \ell _{n_{k}+1}\)  for every \(k\) .
Without loss of generality we
assume that the \((n_k)\)  is strictly increasing and \(n_k\)  is even for every \(k\) .
The proof in the other case is similar.
Let \(e_k=(i_k-\ell _{n_k})/i_k\) .
By passing to a subsequence we may assume that \(e_k\rightarrow e\in [0,1]\)  as \(k\rightarrow \infty \) . From
(REF ) we have that the sequence \(\operatorname{\mathcal {E}}(x, i_k)\)  and
the sequence with \(k\) -th term given by
\(\nu _k=(1-e) \operatorname{\mathcal {E}}(x_{[\ell _{n_{k}-1},\ell _{n_k})})+ e\operatorname{\mathcal {E}}(x_{[\ell _{n_{k}},\ell _{n_k+1})})\) 
has the same weak\({}^*\)  limit in \(\mathcal {M}_T(X)\) .
If \(e=1\) , then \(\nu =\delta _{\bar{0}}\) , so assume \(e<1\) .
Take the sequence \((m_k)\)  so that \(n_k=2m_k\)  for every \(k\) .
Note that \(x_{[\ell _{n_{k}},\ell _{n_k+1})}\)  is a sequence containing only zeros. Also,
\( x_{[\ell _{n_{k}-1},\ell _{n_k})}={\left\lbrace \begin{array}{ll}q_k^{a_{n_k}b_{n_k}}, \quad & \textrm { where } q_k=q(m_k),\textrm { if } t^{(m_k)}\notin \mathcal {T}, \\(y^s_{[0, a_{n_k})})^{b_{n_k}}, \quad & \textrm { where } s = t(m_k), \textrm { if } t^{(m_k)} \in \mathcal {T}.\end{array}\right.}\) 
It follows that the sequence \(\nu _k\)  converges to the same weak* limit as the sequence of measures
\( w_k ={\left\lbrace \begin{array}{ll}(1 - e) \delta _{\overline{q_k}} + e \delta _{\bar{0}}, \quad & \textrm { if } t^{(m_k)}\notin \mathcal {T}, \\(1 - e) \mu _{t^{(m_k)}} + e \delta _{\bar{0}}, & \textrm { if } t^{(m_k)} \in \mathcal {T}.\end{array}\right.}\) 
Suppose \(t^{(m_k)}\notin \mathcal {T}\)  for some subsequence \((k(j))_{j \ge 1}\) . Then the sequence of measures \(\delta _{\overline{q_{k(j)}}}\)  would also converge - contradiction.
Thus for sufficiently big \(k\) , the second case of (REF ) holds. It follows that the sequence
\((\mu _{t^{(m_k)}})\)  converges.
This, by our previous
observations, gives that the \(t^{(m_k)}\)  converge to some \(z \in \omega ^\omega \) .
Since each \(t^{(m_k)} \in \mathcal {T}\) , this show that \(z\)  is a branch through \(\mathcal {T}\) 
and so \(\mathcal {T}\)  is illfounded. We have shown that if \(V(\pi (\mathcal {T})) \ne \lbrace  \delta _{\bar{0}}\rbrace \) ,
then \(\mathcal {T}\)  is illfounded. Conversely, if \(\mathcal {T}\)  is illfounded,
say \(z\)  is a branch through \(\mathcal {T}\) , then along the sequence \((n_{2k})\)  where
\(t^{(k)}=z\upharpoonright k\)  we have that \(\operatorname{\mathcal {E}}(\pi (\mathcal {T}), n_{2k})\)  converges to the measure \(\mu _z\) 
defined above, and \(\mu _z \ne \delta _{\bar{0}}\) , and so we have \(V(\pi (\mathcal {T}))\ne \lbrace  \delta _{\bar{0}}\rbrace \) . So \(V(\pi (\mathcal {T}))=\lbrace \delta _{\bar{0}}\rbrace \) 
if and only if \(\mathcal {T}\in \text{Wf}\) . This finishes the proof. It is easy to see that the above proof shows in fact that \(G_{\lbrace \delta _{\bar{0}}\rbrace }\)  is \(\Pi ^1_1\) -complete, while \({U}{}{G}\) , where \(U=\operatorname{\mathcal {M}_{\mathit {T}} }(X)\setminus \lbrace \delta _{\bar{0}}\rbrace \)  is \(\Sigma ^1_1\)  complete.
